# Prime index subgroup of $\langle Q^{i}(\mathbb Z^{2}) \mid i \in \mathbb Z \rangle$ that is invariant under matrix $Q$

Let $$Q$$ be a matrix in $$\operatorname{GL}(2, \mathbb{Q})$$ and consider the group $$G = \langle Q^{i}(\mathbb Z^{2}) \mid i \in \mathbb Z \rangle := \langle Q^{i}(v) \mid i \in \mathbb Z, v \in \mathbb Z^{2} \rangle$$ under vector addition.

Question: What are the prime index subgroups $$H$$ of $$\langle Q^{i}(\mathbb Z^{2}) \mid i \in \mathbb Z \rangle$$ such that for all $$v \in H$$, we have $$Q(v) \in H$$?

I am interested in the case when both $$Q$$ and $$Q^{-1}$$ have non-integral coefficients in their characteristic polynomial (Here are some equivalent conditions).

"Easy" case: When $$Q$$ (or $$Q^{-1}$$) is an integer matrix we can start with a subgroup $$H'$$ of $$\mathbb Z^2$$ that has finite index $$p$$ and is invariant under $$Q$$, then $$H=\langle Q^{i}( H') \mid i \in \mathbb Z \rangle$$ would be a subgroup of $$G$$ which has finite index $$p$$ and is invariant under $$Q$$.

One of the reasons we can do this is because when $$Q$$ is an integer matrix, we have $$\langle Q^{i}(v)\mid i\in\mathbb Z,v\in\mathbb Z^{2}\rangle = \langle Q^{-i}(v)\mid i\in\mathbb N,v\in\mathbb Z^{2}\rangle$$. However, it is quite difficult to show that this holds for the general rational matrices.

Where I get this question from: I am trying to find all the finite index subgroups of $$G$$ that are invariant under $$Q$$. Given a prime $$p$$ that is coprime to all the denominators of the entries in $$Q$$ and $$Q^{-1}$$, we know that $$\langle Q^{i}(p\mathbb Z^{2}) \mid i \in \mathbb Z \rangle$$ would have finite index $$p^2$$ and is invariant under $$Q$$, I was wondering if there are any other finite index invariant subgroup.

Thank you for reading. Any ideas for constructing such a subgroup would be really appreciated.

• Interesting. I don't have any intuition about this, but to build it from the ground up I am tempted to look at what happens in cases like $$Q=\pmatrix{2&0\cr0&1/3\cr}$$ and $$Q=\pmatrix{0&2\cr 1/3&0\cr}.$$ I don't expect to learn everything from such examples, but may be something. A possible problem may be that Smith normal forms of (suitably scaled) powers of $Q$ don't line up nicely. Scratching my head. Gotta go, sorry. Commented Aug 27, 2023 at 17:49
• Thanks @JyrkiLahtonen. Could you give me a hint of how Smith normal form could help? The span of the column space will change after performing elementary row operations. (Maybe you mean Reduced Column echelon form?) Commented Sep 6, 2023 at 9:43
• May be there is nothing to it? Smith normal form may help for a single matrix in the sense that if $PAQ=D$ with $P,Q$ invertible, then $AQ(\Bbb{Z}^2)=A(\Bbb{Z}^2)$, so $A(\Bbb{Z}^2)=P^{-1}D(\Bbb{Z}^2)$. Yes, the application of $P^{-1}$ will change the group. And you also have the powers $A^i$ to deal with. Commented Sep 6, 2023 at 10:23
• Sorry, you had $Q$ in place of $A$ everywhere. I was too used to using $Q$ as the column operation factor in SNF :-9 Commented Sep 6, 2023 at 10:24